How do you find the vector perpendicular to the plane containing (0,-2,2), (1,2,-3), and (4,0,-1)?

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Answer 1 · 2016-10-20T22:20:36

Make two vectors from the 3 points and then take the cross product:

$barv xx baru = 2hati + 17hatj + 14hatk$

Use the 3 points to create two vectors in the plane:

let $baru = (1 - 0)hati + (2 - -2)hatj + (-3 - 2)hatk$

$baru = 1hati + 4hatj - 5hatk$

let $barv = (4 - 0)hati + (0 - -2)hatj + (-1 - 2)hatk$

$barv = 4hati + 2hatj - 3hatk$

Use the cross product to find a vector perpendicular to these two vectors:

$baru xx barv = |(hati,hatj,hatk,hati,hatj),(1,4,-5,1,4),(4,2,-3,4,2)|$

$baru xx barv = hati{(4)(-3) - (-5)(-2)} + hatj{(-5)(4) - (1)(-3)} + hatk{(1)(2) - (4)(4)}$

$baru xx barv = -2hati - 17hatj - 14hatk$

The negative of this is also valid:

$barv xx baru = 2hati + 17hatj + 14hatk$